Python从单链表中删除节点并输出修改的SLL

Question

我已经编写了一些代码，其中包含了SLL数据结构中的玩家值和他们的分数。所有的工作都很好，直到我试图通过输入一个玩家ID值来删除一个节点，然后输出修改后的SLL而没有玩家得分。你知道吗

下面是我的班级：

class PlayerScore:                          

def __init__(self, score, next = None): 
    self.length=len(score)-1
    self.rep=self.__str(score)
    self.score = score
    self.next = next

def __str(self, score):
    terms = ["(No: "+str(score[0])+ \
             ", Game1: "+str(score[1])+ \
             ", Game2: "+str(score[2]) + \
            ", Game3: "+str(score[3])]
    return str(terms)
def __eq__(self, that):
    return self.score[0] == that[0]

def delete_node(self, data):
    curr = self
    curr_score = curr.score
    prev = None;
    while curr is not None:
        if curr_score == data:
            if prev is not None:
                prev.next = curr.next
            else:
                self = curr.next
        prev = curr
        curr = self.next
    modi = curr_score
    while(modi):
        lyst = modi.marks
        total = lyst[1]+lyst[2] +lyst[3]
        print(" Student_ID.: " + str(lyst[0])+"  A1: " + str(lyst[1])+"  A2: "
            + str(lyst[2])+"  Exam: " + str(lyst[3])+" ->total " + str(total))
        modi = modi.next

主体代码：

def print_score(score_list):
if score_list is None:
    print("No player records")
    return
else:
    print("Original linked list:")
    curr = score_list
    while curr is not None:
        lyst = curr.score
        print(" PlayerNo: " + str(lyst[0])+"  Game 1: " + str(lyst[1])+"  Game 2: "
              + str(lyst[2])+"  Game 3: " + str(lyst[3]))
        curr = curr.next
    print()

def main(size = 4):

    node1 = None

    #create SLL    
    node2 = PlayerScore([199, 94, 96, 109], node1)
    node3 = PlayerScore([185, 203,  156, 171], node2)
    node4 = PlayerScore([173, 104, 190, 224], node3)
    node5 = PlayerScore([154, 268,  287, 300], node4)

    player_score_head = node5

    print_score(player_score_head)

    value = input("Enter a Student's ID for deletion: ")
    print(' ')
    player_score_head.delete_node(value)

if __name__ == "__main__":
    main() 

我的问题是我在delete_node内哪里出错了？我确实认为，代码是工作，直到点，它必须打印新的SLL，但我不能在哪里锻炼。也许我忽略了什么，只是需要一双新的眼睛来指出我愚蠢的错误。你知道吗

请注意，我对数据结构和算法的概念还比较陌生，所以如果我做错了什么或者这个问题没有意义，请原谅我的无知。你知道吗

感谢您的帮助和反馈。你知道吗

Answer 1

这是一个带有delete_node方法的代码修改版本，我对PlayerScore类做了一些其他更改。我将ID号从分数数据中分离出来；ID号不是分数，所以将它们结合起来是没有意义的。我给了PlayerScore一个__str__方法，使打印节点更容易。我摆脱了__eq__方法，因为我发现它使代码更难阅读和分析。你知道吗

我从delete_node中删除了modi内容；这些内容与节点删除无关，因此它属于一个单独的方法。你知道吗

我还对main做了一些更改。我给了它一个输入循环，以便我们可以尝试删除几个节点来测试delete_node方法。我添加了一个测试来验证输入的ID号是否可以转换为整数。你知道吗

class PlayerScore:
    def __init__(self, data, nxt=None):
        self.idnum, *self.scores = data
        self.nxt = nxt

    def __len__(self):
        return len(self.scores)

    def __str__(self):
        terms = ["No: " + str(self.idnum)]
        terms += ["Game {}: {}".format(i, v) 
            for i, v in enumerate(self.scores, 1)]
        return ', '.join(terms)

    def delete_node(self, idnum):
        curr = self
        prev = None
        # Find the node with idnum
        while curr is not None:
            if curr.idnum == idnum:
                break
            prev = curr
            curr = curr.nxt
        else:
            print("Node {} not found".format(idnum))
            return self

        #print('CURR', curr, 'PREV', prev)

        if prev is None:
            new_head = curr.nxt
        else:
            new_head = self
            prev.nxt = curr.nxt

        del curr
        return new_head

def print_score(score_list):
    if score_list is None:
        print("No player records")
        return

    print("Original linked list:")
    curr = score_list
    while curr is not None:
        print(curr)
        curr = curr.nxt
    print()

def main():
    data_list = [
        [199, 94, 96, 109],
        [185, 203,  156, 171],
        [173, 104, 190, 224],
        [154, 268,  287, 300],
    ]

    #create SLL
    head = None
    for data in data_list:
        head = PlayerScore(data, head)

    print_score(head)

    while head is not None:
        value = input("Enter a Student's ID for deletion, or 0 to exit: ")
        try:
            value = int(value)
        except ValueError:
            print("ID must be an integer")
            continue
        if value == 0:
            break
        head = head.delete_node(value)
        print_score(head)


if __name__ == "__main__":
    main()

演示输出

Original linked list:
No: 154, Game 1: 268, Game 2: 287, Game 3: 300
No: 173, Game 1: 104, Game 2: 190, Game 3: 224
No: 185, Game 1: 203, Game 2: 156, Game 3: 171
No: 199, Game 1: 94, Game 2: 96, Game 3: 109

Enter a Student's ID for deletion, or 0 to exit: abc
ID must be an integer
Enter a Student's ID for deletion, or 0 to exit: 200
Node 200 not found
Original linked list:
No: 154, Game 1: 268, Game 2: 287, Game 3: 300
No: 173, Game 1: 104, Game 2: 190, Game 3: 224
No: 185, Game 1: 203, Game 2: 156, Game 3: 171
No: 199, Game 1: 94, Game 2: 96, Game 3: 109

Enter a Student's ID for deletion, or 0 to exit: 199
Original linked list:
No: 154, Game 1: 268, Game 2: 287, Game 3: 300
No: 173, Game 1: 104, Game 2: 190, Game 3: 224
No: 185, Game 1: 203, Game 2: 156, Game 3: 171

Enter a Student's ID for deletion, or 0 to exit: 154
Original linked list:
No: 173, Game 1: 104, Game 2: 190, Game 3: 224
No: 185, Game 1: 203, Game 2: 156, Game 3: 171

Enter a Student's ID for deletion, or 0 to exit: 173
Original linked list:
No: 185, Game 1: 203, Game 2: 156, Game 3: 171

Enter a Student's ID for deletion, or 0 to exit: 185
No player records